chemistry 102(01) spring 2009
DESCRIPTION
Chemistry 102(01) spring 2009. Instructor: Dr. Upali Siriwardane e-mail : [email protected] Office : CTH 311 Phone 257-4941 Office Hours : M,W, 8:00-9:00 & 11:00-12:00 a.m.; Tu,Th,F 9:00 - 10:00 a.m. - PowerPoint PPT PresentationTRANSCRIPT
Instructor: Dr. Upali Siriwardanee-mail: [email protected]
Office: CTH 311 Phone 257-4941
Office Hours: M,W, 8:00-9:00 & 11:00-12:00 a.m.; Tu,Th,F 9:00 - 10:00 a.m.
Test Dates: March 25, April 26, and May 18; Comprehensive Final Exam: May 20,2009 9:30-10:45
am, CTH 328.March 30, 2009 (Test 1): Chapter 13 April 27, 2009 (Test 2): Chapters 14 & 15 May 18, 2009 (Test 3): Chapters 16, 17 & 18
Comprehensive Final Exam: May 20,2009 :Chapters 13, 14, 15, 16, 17 and 18
Chemistry 102(01) spring 2009
Review of Chapter 6. Energy and Chemical Review of Chapter 6. Energy and Chemical ReactionsReactions
6.1 The Nature of Energy
6.2 Conservation of Energy
6.3 Heat Capacity
6.4 Energy and Enthalpy
6.5 Thermochemical Equations
6.6 Enthalpy change for chemical Rections
6.7 Where does the Energy come from?
6.8 Measuring Enthalpy Changes: Calorimetry
6.9 Hess's Law
6.10 Standard Enthalpy of Formation
6.11 Chemical Fuels for Home and Industry
6.12 Food Fuels for Our Bodies
ThermochemistryThermochemistry
Heat changes during chemical reactions
Thermochemical equation. eg.
H2 (g) + O2 (g) ---> 2H2O(l) H =- 256 kJ;
is called the enthalpy of reaction.
if H is + reaction is called endothermic
if H is - reaction is called exothermic
Why is it necessary to divide Why is it necessary to divide Universe into System and Universe into System and
SurroundingSurroundingUniverse = System + Universe = System +
SurroundingSurrounding
system surroundings
universe
Types of SystemsTypes of SystemsIsolated systemIsolated system
no mass or energy exchange
Closed systemClosed systemonly energy exchange
Open systemOpen systemboth mass and energy exchange
Universe = System + SurroundingUniverse = System + Surrounding
Why is it necessary to divide Universe into Why is it necessary to divide Universe into System and SurroundingSystem and Surrounding
What is the internal energy change (What is the internal energy change (U) U) of a system?of a system?
U is associated with changes in atoms, molecules U is associated with changes in atoms, molecules and subatomic particlesand subatomic particles
Etotal = Eke + E pe + U
U = heat (q) + w (work)
U = q + w
U = q -P V; w =- P V
What forms of energy are found in What forms of energy are found in the Universe?the Universe?
mechanical mechanical
thermalthermal
electrical electrical
nuclearnuclear
mass: E = mcmass: E = mc22
others yet to discoverothers yet to discover
What is 1What is 1stst Law of Thermodynamics Law of Thermodynamics
Eenergy is conserved in the UniverseEenergy is conserved in the Universe
All forms of energy are inter-convertible All forms of energy are inter-convertible and conservedand conserved
Energy is neither created nor destroyed.Energy is neither created nor destroyed.
What exactly is What exactly is H?H?
Heat measured at constant pressure qp
Chemical reactions exposed to atmosphere and are held at a constant pressure.
Volume of materials or gases produced can change.
Volume expansion work = -PV
U = qp + w; U = qp -PV
qp = U + PV; w = -PV
H = U + PV; qp = H(enthalpy )
Heat measured at constant volume qv
Chemical reactions take place inside a bomb. Volume of materials or gases produced can not change. ie: work = -PV= 0 U = qv + wqv = U + o; w = 0U = qv = U(internal energy )
How do you measure How do you measure U?U?
What is Hess's Law of Summation of What is Hess's Law of Summation of Heat?Heat?
To heat of reaction for new reactions.
Two methodsTwo methods?
1st method: new H is calculated by adding Hs of other reactions.
2nd method2nd method: Where Hf ( H of formation) of reactants and products are used to calculate H of a reaction.
Method 1: Calculate Method 1: Calculate H for the reaction:H for the reaction:
SO2(g) + 1/2 O2(g) + H2O(g) ----> H2SO4(l) H = ?
Other reactions:
SO2(g) ------> S(s) + O2(g) ; H = 297kJ
H2SO4(l)------> H2(g) + S(s) + 2O2(g); H = 814 kJ
H2(g) +1/2O2(g) -----> H2O(g); H = -242 kJ
SO2(g) ------> S(s) + O2(g); H1 = 297 kJ - 1
H2(g) + S(s) + 2O2(g) ------> H2SO4(l); H2 = -814 kJ - 2
H2O(g) ----->H2(g) + 1/2 O2(g) ; H3 = +242 kJ - 3
______________________________________
SO2(g) + 1/2 O2(g) + H2O(g) -----> H2SO4(l); H = ?
H = H1+ H2+ H3
H = +297 - 814 + 242
H = -275 kJ
Calculate Calculate H for the reactionH for the reaction
Calculate Heat (Enthalpy) of Calculate Heat (Enthalpy) of Combustion: 2nd methodCombustion: 2nd method
C7H16(l) + 11 O2(g) -----> 7 CO2(g) + 8 H2O(l) ; Ho = ?
Hf (C7H16) = -198.8 kJ/mol
Hf (CO2) = -393.5 kJ/mol
Hf (H2O) = -285.9 kJ/mol
Hf O2(g) = 0 (zero)
What method?Ho = n Hf
o products – n Hfo reactants
n = stoichiometric coefficients
2nd method
H = [n ( Hof) Products] - [n (Ho
f) reactants]
H = [ 7(- 393.5 + 8 (- 285.9)] - [-198.8 + 11 (0)]
= [-2754.5 - 2287.2] - [-198.8]
= -5041.7 + 198.8
= -4842.9 kJ = -4843 kJ
Calculate Calculate H for the reactionH for the reaction
Why is Why is HHoof f of elements is of elements is zero?zero?
Hof, Heat formations are for compounds
Note: Hof of elements is zero
Chapter 18. Thermodynamics: Directionality of Chemical Reactions
18.1 Reactant-Favored and Product-Favored Processes
18.2 Probability and Chemical Reactions18.3 Measuring Dispersal or Disorder: Entropy18.4 Calculating Entropy Changes18.5 Entropy and the Second Law of
Thermodynamics18.6 Gibbs Free Energy18.7 Gibbs Free Energy Changes and Equilibrium Constants18.8 Gibbs Free Energy, Maximum Work, and
Energy Resources18.9 Gibbs Free Energy and Biological Systems18.10 Conservation of Gibbs Free Energy18.11 Thermodynamic and Kinetic Stability
Chemical ThermodynamicsChemical Thermodynamics
spontaneous reaction – reaction which proceed without external assistance once started
chemical thermodynamics helps predict which reactions are spontaneous
ThermodynamicsThermodynamicsThermodynamicsThermodynamics
Will the rearrangement of a system decrease its Will the rearrangement of a system decrease its energy? energy?
If yes, system is favored to react — a If yes, system is favored to react — a product-product-favoredfavored system.system.
Most product-favored reactions are Most product-favored reactions are exothermic.exothermic.
Often referred to as Often referred to as spontaneousspontaneous reactions.reactions.
““Spontaneous” does not imply anything about time Spontaneous” does not imply anything about time for reaction to occur. Kinetic factors are more for reaction to occur. Kinetic factors are more important for certain reactions.important for certain reactions.
Thermodynamics and KineticsThermodynamics and KineticsThermodynamics and KineticsThermodynamics and KineticsDiamond Diamond graphite graphite
Thermodynamically Thermodynamically product favoredproduct favored
Slow KineticsSlow Kinetics
Paper burns
Thermodynamically Thermodynamically product favoredproduct favored
Fast KineticsFast Kinetics
In this chapter we only In this chapter we only look into look into thermodynamic thermodynamic factorsfactors
Bases on Energy: Product-Bases on Energy: Product-Favored ReactionsFavored Reactions
In general, product-favored reactions are In general, product-favored reactions are exothermicexothermic..
(Negative H) H)
In general, reactant-favored reactions are In general, reactant-favored reactions are endothermicendothermic..
(Positive H)H)
Product-Favored ReactionsProduct-Favored ReactionsBut many spontaneous reactions or processes But many spontaneous reactions or processes
are endothermic or even have are endothermic or even have H = 0.H = 0.
NHNH44NONO33(s) (s) NH NH44NONO33(aq); (aq); H = +H = +
Direction of ReactionDirection of ReactionProduct favored reactions are always a
transformation of a reactants favored reaction.
Product Favored ReactionProduct Favored Reaction
2Na(s) + 2Cl2(g) => 2NaCl(s)
Reactant Favored ReactionReactant Favored Reaction
2NaCl(s) => 2Na(s) + 2Cl2(g)
However, non spontaneous reactions could be made to take However, non spontaneous reactions could be made to take place by coupling with energy source: another reaction or place by coupling with energy source: another reaction or electric currentelectric current
ElectrolysisElectrolysis
2NaCl(l) => 2Na(s) + 2Cl2(g)
Expansion of a Gas
The positional probability is higher when particles are dispersed over a larger volume
Matter tends to expand unless it is restricted
Entropy, SEntropy, SEntropy, SEntropy, SThe thermodynamic property The thermodynamic property
related to randomness is related to randomness is ENTROPY, SENTROPY, S..
Product-favored processes: Product-favored processes: final state is more final state is more DISORDEREDDISORDERED or or RANDOMRANDOM than the original.than the original.
Spontaneity is related to Spontaneity is related to an increase in an increase in randomness.randomness.
Reaction of K Reaction of K with waterwith water
Entropies of Solid, Liquidand Gas Phases
S (gases) > S (liquids) > S (solids)S (gases) > S (liquids) > S (solids)
Entropy, SEntropy, SEntropy, SEntropy, S
Entropies of ionic solids depend on Entropies of ionic solids depend on coulombic attractions.coulombic attractions.
SSoo (J/K•mol) (J/K•mol)
MgOMgO 26.926.9
NaFNaF 51.551.5
SSoo (J/K•mol) (J/K•mol)
MgOMgO 26.926.9
NaFNaF 51.551.5
Qualitative Guidelines for Entropy Changes
Entropies of gases higher than liquids higher than solids
Entropies are higher for more complex structures than simpler structures
Entropies of ionic solids are inversely related to the strength of ionic forces
Entropy increases when making solutions of pure solids or pure liquids in a liquid solvent
Entropy decrease when making solutions of gases in a liquid
Phase TransitionsPhase Transitions
H2O(s) => H2O(l) H > 0; S > 0
H2O(l) => H2O(g) H > 0; S > 0
spontaneous at high temperatures
H2O(l) => H2O(s) H < 0; S < 0
H2O(g) => H2O(l) H < 0; S < 0
spontaneous at low temperatures
Entropy Changes for Phase Entropy Changes for Phase ChangesChanges
For a phase change, For a phase change, SSSYSSYS = q = qSYSSYS/T/T
(q = heat transferred)(q = heat transferred)
Boiling WaterBoiling Water
HH22O (liq) O (liq) H H22O(g)O(g)
H = q = +40,700 J/molH = q = +40,700 J/mol
mol•J/K 109+ = K 373.15
J/mol 40,700 =
T
q = S
Phase TransitionsPhase Transitions
Heat of Fusion
energy associated with phase transition solid-to-liquid or liquid-to-solid
Gfusion = 0 = Hfusion - T Sfusion
0 = Hfusion - T Sfusion
Hfusion = T Sfusion
Heat of Vaporization
energy associated with phase transition gas-to-liquid or liquid-to-gas
Hvaporization = T Svaporization
Qualitative prediction of Qualitative prediction of S of S of Chemical ReactionsChemical Reactions
• Look for (l) or (s) --> (g)
• If all are gases: calculate nn = n (gaseous prod.) - n(gaseous reac.)
N2 (g) + 3 H2 (g) --------> 2 NH3 (g)
n = 2 - 4 = -2
If n is - S is negative (decrease in S)
If n is + S is positive (increase in S)
Entropy ChangeEntropy Change
Entropy (Entropy (S) normally increase (+) for the following S) normally increase (+) for the following changes:changes:
i) Solid ---> liquid (melting) ++
ii) Liquid ---> gas ++
iii) Solid ----> gas most +most +
iv) Increase in temperature ++
v) Increasing in pressure(constant volume, and temperature) ++
vi) Increase in volume ++
Predict Predict S!S!
2 C2H6(g) + 7 O2(g)--> 4 CO2(g) + 6H2O(g)
2 CO(g) + O2(g)-->2 CO2(g)
HCl(g) + NH3(g)-->NH4Cl(s)
H2(g) + Br2(l) --> 2 HBr(g)
2 H2 H22(g) + O(g) + O22(g) (g) 2 H 2 H22O(liq)O(liq)
SSoo = 2 S = 2 Soo (H (H22O) - [2 SO) - [2 Soo (H (H22) + S) + Soo (O (O22)])]
SSoo = 2 mol (69.9 J/K•mol) – = 2 mol (69.9 J/K•mol) – [2 mol (130.7 J/K•mol) + 1 mol (205.3 J/K•mol)][2 mol (130.7 J/K•mol) + 1 mol (205.3 J/K•mol)]
SSoo = -326.9 J/K = -326.9 J/K
There is a There is a decrease in S decrease in S because 3 mol of gas give because 3 mol of gas give 2 mol of liquid.2 mol of liquid.
Calculating Calculating S for a ReactionS for a Reaction Based on Hess’s Law second method:
SSoo = = S Soo (products) - (products) - S Soo (reactants) (reactants)
Based on Hess’s Law second method:
SSoo = = S Soo (products) - (products) - S Soo (reactants) (reactants)
Third Law of ThermodynamicsThird Law of Thermodynamics
Provides reference point for absolute entropy
Entropy of a perfectly crystalline substance at absolute zero (T= 0 K) is zero.
Unlike H entropy values are positive above temperatures above absolute zero.
Substance So (J/K.mol) Substance So (J/K.mol)
C (diamond) 2.37 HBr (g) 198.59
C (graphite) 5.69 HCl (g) 186.80
CaO (s) 39.75 HF (g) 193.67
CaCO3 (s) 92.9 HI (g) 206.33
C2H2 (g) ` 200.82 H2O (l) 69.91
C2H4 (g) 219.4 H2O (g) 188.72
C2H6 (g) 229.5 NaCl (s) 72.12
CH3OH (l) 127 O2 (g) 205.03
CH3OH (g) 238 SO2 (g) 248.12
CO (g) 197.91 SO3 (g) 256.72
Standard Entropies at 25Standard Entropies at 25ooCC
Entropy & SpontaneityEntropy & Spontaneity
• How can water boil and freeze spontaneously?
• Enthalpy change can not predict spontaneity!
• Some endothermic processes are spontaneous
• Need another thermodynamic property.
Laws of ThermodynamicsLaws of Thermodynamics
FirstFirst : The total energy of the universe is : The total energy of the universe is constantconstant
SecondSecond : The total entropy (S) of the : The total entropy (S) of the universe is always increasinguniverse is always increasing
Third Third : The entropy(S) of a pure, : The entropy(S) of a pure, perfectly formed crystalline substance at perfectly formed crystalline substance at absolute zero is zeroabsolute zero is zero
FirstFirst : The total energy of the universe is : The total energy of the universe is constantconstant
SecondSecond : The total entropy (S) of the : The total entropy (S) of the universe is always increasinguniverse is always increasing
Third Third : The entropy(S) of a pure, : The entropy(S) of a pure, perfectly formed crystalline substance at perfectly formed crystalline substance at absolute zero is zeroabsolute zero is zero
Dissolving NH4NO3 in water—an entropy driven process.
NHNH44NONO33(s) (s) NH NH44NONO33(aq); (aq); H = +H = +
2nd Law of Thermodynamics2nd Law of Thermodynamics
Second Law of ThermodynamicsSecond Law of ThermodynamicsIn the universe the ENTROPY cannot decrease for
any spontaneous process
The entropy of the universe strives for a maximum
in any spontaneous process, the entropy of the universe increases
for product-favored processfor product-favored process
Suniverse = ( Ssys + Ssurr) > 0
Suniv = entropy of the Universe
Ssys = entropy of the System
Ssurr = entropy of the Surrounding
Suniv = Ssys + Ssurr
Entropy of the UniverseEntropy of the Universe
Suniv = Ssys + Ssurr
Suniv Ssys Ssurr
+ + +
+ +(Ssys>Ssurr) -
+ - + (Ssurr>Ssys)
Can calc. that Can calc. that HHoorxnrxn = = HHoo
systemsystem = -571.7 kJ = -571.7 kJCan calc. that Can calc. that HHoorxnrxn = = HHoo
systemsystem = -571.7 kJ = -571.7 kJ
2 H2 H22(g) + O(g) + O22(g) (g) 2 H 2 H22O(liq)O(liq)
SSoosyssys = -326.9 J/K = -326.9 J/K
Entropy Changes in the Surroundings
T
H- =
T
q = systemsurr
gssurroundin
oS T
H- =
T
q = systemsurr
gssurroundin
oS
K 298.15
J/kJ) kJ)(1000 (-571.7 - = gssurroundin
oS K 298.15
J/kJ) kJ)(1000 (-571.7 - = gssurroundin
oS
= = +1917 J/K+1917 J/K= = +1917 J/K+1917 J/K
2nd Law of Thermodynamics2nd Law of Thermodynamics
2 H2 H22(g) + O(g) + O22(g) (g) 2 H 2 H22O(liq)O(liq)
SSoosyssys = -326.9 J/K = -326.9 J/K
SSoosurrsurr = +1917 J/K = +1917 J/K
SSoouni uni = +1590. J/K= +1590. J/K
The entropy of the universe is increasing, so The entropy of the universe is increasing, so the reaction is product-favored. the reaction is product-favored.
2nd Law of Thermodynamics2nd Law of Thermodynamics
Gibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, G
SSunivuniv = = SSsurrsurr + + SSsyssys
Multiply through by (-T)Multiply through by (-T)
-T-TSSunivuniv = = HHsyssys - T - TSSsyssys
-T-TSSunivuniv = = GGsystemsystem
Under standard conditions —Under standard conditions —
GGoo = = HHoo - T - TSSoo
Suniv = Hsys
T + Ssys
Gibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, G
GGoo = = HHoo - T - T SSoo
Gibbs free energy change = Gibbs free energy change = difference between the enthalpy of a system and
the product of its absolute temperature and entropy
predictor of spontaneity
Total energy change for system -Total energy change for system -energy lost in disordering the systemenergy lost in disordering the system
Predict the spontaneity of the Predict the spontaneity of the following processes from following processes from H and H and S S
at various temperatures.at various temperatures.
a)H = 30 kJ, S = 6 kJ, T = 300 Kb)H = 15 kJ,S = -45 kJ,T = 200 K
a)H = 30 kJ S = 6 kJ T = 300 K G = Hsys-TSsys or G = H - TS.
H = 30 kJS = 6 kJ T = 300 K
G = 30 kJ - (300 x 6 kJ) = 30 -1800 kJ
G = -1770 kJ
b) H = 15 kJ S = -45 kJ T = 200 KG = Hsys-TSsys or G = H - TS.
H = 15 kJS = -45 kJ T = 200 K
G = 15 kJ -[200 (-45 kJ)] = 15 kJ -(-9000) kJ
G = 15 + 9000 kJ = 9015 kJ
The sign ofGG indicates whether a reaction will occur spontaneously.
++ Not spontaneousNot spontaneous
0 0 At equilibriumAt equilibrium
-- SpontaneousSpontaneous
The fact that the effect of SS will vary as a function of temperature is important. This can result in changing the sign of GG.
Free energyFree energy
Predicting Whether a ReactionPredicting Whether a Reactionis Product Favored using is Product Favored using GG
Sign of Hsystem Sign of Ssystem Product-favored?
Negative (exothermic) Positive Yes
Negative (exothermic) Negative Yes at low T; no at
high T
Positive (endothermic) Positive No at low T;
yes at high T
Positive (endothermic) Negative No
Predict Predict G at different G at different H, H, S, TS, T
G = G = H - T H - T S.S.
- - all T T +
+ + all T T –
- /+ + high/low TT +
-/+ - low/high T T -
Gibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, G
GGoo = = HHoo - T - TSSoo
HHoo SSoo GGoo ReactionReaction
exo(-)exo(-) increase(+)increase(+) -- Prod-favoredProd-favored
endo(+)endo(+) decrease(-)decrease(-) ++ React-favoredReact-favored
exo(-)exo(-) decrease(-)decrease(-) ?? T dependentT dependent
endo(+)endo(+) increase(+)increase(+) ?? T dependentT dependent
GGffoo
Free energy change that results when one mole of a substance if formed from its elements will all substances in their standard states.
G values can then be calculated from:
GGoo = = nnppGGffoo productsproducts – – nnrrGGff
oo reactantsreactants
Standard free energy of formation, Standard free energy of formation, GGffoo
SubstanceSubstance GGffoo SubstanceSubstance GGff
oo
C (diamond) 2.832 HBr (g) -53.43
CaO (s) -604.04 HF (g) -273.22
CaCO3 (s) -1128.84 HI (g) 1.30
C2H2 (g) 209 H2O (l) -237.18
C2H4 (g) 86.12 H2O (g) -228.59
C2H6 (g) -32.89 NaCl (s) -384.04
CH3OH (l) -166.3 O (g) 231.75
CH3OH (g) -161.9 SO2 (g) -300.19
CO (g) -137.27 SO3 (g) -371.08
All have units of kJ/mol and are for 25 All have units of kJ/mol and are for 25 ooCC
Standard free energy of formationStandard free energy of formation
How do you calculate How do you calculate GG
There are two ways to calculateG
for chemical reactions.
i) G = H - TS.
ii) Go = Gof (products) - G o
f (reactants)
Gibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, G
GGoo = = HHoo - T - TSSoo
Two methods of calculating Two methods of calculating GGoo
(a) Determine (a) Determine HHoorxnrxn and and SSoo
rxnrxn and use Gibbs and use Gibbs
equation.equation.
(b) Use tabulated values of free energies of (b) Use tabulated values of free energies of
formation, formation, GGffoo..
GGoorxnrxn = = GGff
oo (products) - (products) - GGffoo (reactants) (reactants)GGoo
rxnrxn = = GGffoo (products) - (products) - GGff
oo (reactants) (reactants)
Calculating Calculating GGoorxnrxn
Calculating Calculating GGoorxnrxn
Is the dissolution of ammonium nitrate product-Is the dissolution of ammonium nitrate product-favored? favored?
If so, is it enthalpy- or entropy-driven?If so, is it enthalpy- or entropy-driven?
NHNH44NONO33(s) + heat (s) + heat NH NH44NONO33(aq)(aq)
Calculating Calculating GGoorxnrxn
Calculating Calculating GGoorxnrxn
Method (a)Method (a) : From tables of thermodynamic data we find : From tables of thermodynamic data we find
HHoorxnrxn = +25.7 kJ = +25.7 kJ
SSoorxnrxn = +108.7 J/K or +0.1087 kJ/K = +108.7 J/K or +0.1087 kJ/K
GGoorxnrxn = +25.7 kJ - (298 K)(+0.1087 J/K) = +25.7 kJ - (298 K)(+0.1087 J/K)
= -6.7 kJ= -6.7 kJ
Reaction is Reaction is product-favoredproduct-favored in spite of negative in spite of negative HHoorxnrxn. .
Reaction is Reaction is “entropy driven”“entropy driven”
NHNH44NONO33(s) + heat (s) + heat NH NH44NONO33(aq)(aq)
Calculating Calculating GGoorxnrxn
Calculating Calculating GGoorxnrxn
Combustion of carbonCombustion of carbon
C(graphite) + OC(graphite) + O22(g) --> CO(g) --> CO22(g) (g) Method (b) :
GGoorxnrxn = = GGff
oo(CO(CO22) - [) - [GGffoo(graph) + (graph) + GGff
oo(O(O22)])]
GGoorxnrxn = -394.4 kJ - [ 0 + 0] = -394.4 kJ - [ 0 + 0]
Note that free energy of formation of an element in Note that free energy of formation of an element in its standard state is 0.its standard state is 0.
GGoorxnrxn = -394.4 kJ = -394.4 kJ
Reaction is Reaction is product-favoredproduct-favored
GGoorxnrxn = = GGff
oo (products) - (products) - GGffoo (reactants) (reactants)GGoo
rxnrxn = = GGffoo (products) - (products) - GGff
oo (reactants) (reactants)
We can calculate Go values from Ho and So values at a constant temperature and pressure.
Example.Example.
Determine Go for the following reaction at 25oC
Equation N2 (g) + 3H2 (g) 2NH3 (g)
Hfo, kJ/mol 0.00 0.00 -46.11
So, J/K.mol 191.50 130.68 192.3
Calculation of Go
Example.Example.
Calculate the Sorxn at 25 oC for the following
reaction.
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)
SubstanceSubstance SSoo (J/K (J/K..mol)mol)
CH4 (g) 186.2
O2 (g) 205.03
CO2 (g) 213.64
H2O (g) 188.72
Calculation of standard entropy changesCalculation of standard entropy changes
Calculate the S for the following reactions usingSo
= So (products) - S o(reactants)a) 2SO2 (g) + O2 (g) ------> 2SO 3(g)
So [SO2(g)] = 248 J/K mole ; So [O2(g)] = 205 J/K mole; So [SO3(g)] = 257 J/K mole
b) 2NH 3 (g) + 3N2O (g) --------> 4N2 (g) + 3 H2O (l)
So[ NH3(g)] = 193 J/K mole ; So [N2(g)] = 192 J/K mole;
So [N2O(g)] = 220 J/K mole; S[ H2O(l)] = 70 J/K mole
a)a) 2SO2SO22 (g) + O (g) + O22 (g------> 2SO (g------> 2SO 33(g)(g) So [SO2(g)] = 248 J/K mole ; So [O2(g)] = 205 J/K mole; So [SO3(g)] = 257 J/K moleSo 496 205 514
So = So (products) - S o(reactants)
So = [514] - [496 + 205]
So = 514 - 701
So = -187 J/K mole
Calculate the Calculate the G value for the following G value for the following reactions using:reactions using: Go = Go
f (products) - Gof (reactants)
N2O5 (g) + H2O(l) ------> 2 HNO3(l) ; Go = ? Gf
o[ N2O5 (g) ] = 134 kJ/mole ; Gfo [H2O(g)] = -237 kJ/mole;
Gfo[ HNO3(l) ] = -81 kJ/mole
N2O5 (g) + H2O(l) ------> 2 HNO3(l) ; Go = ?Gf
o 1 x 134 1 x (-237) 2 (-81) 134 -237 -162 Go = Go
f (products) - 3 Gof (reactants)
Go = [-162] - [134 + (-237)]Go = -162 + 103Go = -59 kJ/mole The reaction have a negative G and the reaction is spontaneous or will take place as written.
Free Energy and TemperatureFree Energy and Temperature
2 Fe2 Fe22OO33(s) + 3 C(s) ---> 4 Fe(s) + 3 CO(s) + 3 C(s) ---> 4 Fe(s) + 3 CO22(g)(g)
HHoorxnrxn = +467.9 kJ = +467.9 kJ SSoo
rxnrxn = +560.3 J/K = +560.3 J/K
GGoorxnrxn = +300.8 kJ = +300.8 kJ
Reaction is Reaction is reactant-favoredreactant-favored at 298 K at 298 K
At what T does At what T does GGoorxnrxn change from (+) to (-)? change from (+) to (-)?
Set Set GGoorxnrxn = 0 = = 0 = HHoo
rxnrxn - T - TSSoorxnrxn
K 835.1 = kJ/K 0.5603
kJ 467.9 =
S
H = T
rxn
rxn
How do you calculate How do you calculate G at G at different T and Pdifferent T and P
G = Go + RT ln Q
Q = reaction quotient
at equilibrium G = = Go + RT ln K
Go = - RT ln KIf you know Go you could calculate K
Concentrations, Free Energy, and Concentrations, Free Energy, and the Equilibrium Constantthe Equilibrium Constant
Equilibrium Constant and Free Energy
G = Go + RT ln Q
Q = reaction quotient
0 = Go + RT ln Keq
Go = - RT ln Keq
KKeqeq is related to reaction favorability and so to is related to reaction favorability and so to GGoorxnrxn..
The larger the (-) value of The larger the (-) value of GGoorxnrxn the larger the value the larger the value
of K.of K.
GGoorxnrxn = - RT lnK = - RT lnK
where R = 8.31 J/K•molwhere R = 8.31 J/K•mol
Thermodynamics and KThermodynamics and Keqeq
For gases, the equilibrium constant for a reaction can be related to Go by:
GGoo = - = -RTRT ln lnKK
For our earlier example,
N2 (g) + 3H2 (g) 2NH3 (g)
At 25oC, Go was -32.91 kJ so K would be:
ln K = =
ln K = 13.27; K = 5.8 x 105
Go
-RT-32.91 kJ
-(0.008315 kJ.K-1mol-1)(298.2K)
Free energy and equilibriumFree energy and equilibrium
Calculate the Calculate the G for the following G for the following equilibrium reaction and predict the equilibrium reaction and predict the direction of the change using the equation:direction of the change using the equation: G = Go + RT ln Q ; [Gf
o[ NH3(g) ] = -17 kJ/mole
N2 (g) + 3 H2 (g) 2 NH3 (g); G = ? at 300 K, PN2 = 300, PNH3 = 75 and PH2 = 300
N2 (g) + 3 H2 (g) 2 NH3 (g); G = ?
To calculate To calculate GGoo
Using Go = Gof (products) - Go
f (reactants)
Gfo[ N2(g) ] = 0 kJ/mole; Gf
o[ H2(g) ] = 0
kJ/mole; Gfo[ NH3(g) ] = -17 kJ/mole
Notice elements have Gfo = 0.00 similar to Hf
o
N2 (g) + 3 H2 (g) 2 NH3 (g); G = ?Gf
o 0 0 2 x (-17) 0 0 -34 Go = Go
f (products) - Gof (reactants)
Go = [-34] - [0 +0]Go = -34Go = -34 kJ/mole
To calculate QTo calculate QEquilibrium expression for the reaction in terms of partial pressure:N2 (g) + 3 H2 (g) 2 NH3 (g) p2
NH3
K = _________ pN2 p3
H2
p2NH3
Q = _________ ; pN2 p3
H2
Q is when initial concentration is substituted into the equilibrium expression 752
Q = _________ ; p2NH3= 752; pN2 =300; p3
H2=3003
300 x 3003
Q = 6.94 x 10-7
To calculate To calculate GGoo
G = Go + RT ln Q
Go= -34 kJ/mole
R = 8.314 J/K mole or 8.314 x 10-3kJ/Kmole
T = 300 K
Q= 6.94 x 10-7
G = (-34 kJ/mole) + ( 8.314 x 10-3 kJ/K mole) (300 K) ( ln
6.94 x 10-7)
G = -34 + 2.49 ln 6.94 x 10-7
G = -34 + 2.49 x (-14.18)
G = -34 -35.37
G = -69.37 kJ/mole
Calculate K (from Calculate K (from GG00))
NN22OO44 --->2 NO --->2 NO22 GGoorxnrxn = +4.8 kJ = +4.8 kJ
GGoorxnrxn = +4800 J = - (8.31 J/K)(298 K) ln K = +4800 J = - (8.31 J/K)(298 K) ln K
GGoorxnrxn = - RT lnK = - RT lnK
1.94- = K)J/K)(298 (8.31
J 4800 - = lnK
K = 0.14K = 0.14GGoo
rxnrxn > 0 : K < 1 > 0 : K < 1GGoo
rxnrxn < 0 : K > 1 < 0 : K > 1
K = 0.14K = 0.14GGoo
rxnrxn > 0 : K < 1 > 0 : K < 1GGoo
rxnrxn < 0 : K > 1 < 0 : K > 1
Thermodynamics and KThermodynamics and Keqeq
Concentrations, Free Energy, and the Equilibrium Constant
The Influence of Temperature on Vapor Pressure
H2O(l) => H2O(g)
Keq = pwater vapor
pwater vapor = Keq = e- G'/RT
G as a Function of theG as a Function of theExtent of the ReactionExtent of the Reactionwhen there is Mixingwhen there is Mixing
Coupled ReactionsCoupled Reactions
How to do a reaction that is not thermodynamically favorable?
Find a reaction that offset the (+) G
Thermite ReactionThermite Reaction
Fe2O3(s) => 2Fe(s) + 3/2O2(g)
2Al(s) + 3/2O2(g) Al2O3(s)
Using Electricity for reactions with (+) G: Electrolysis
Non spontaneous reactions could be Non spontaneous reactions could be made to take place by coupling with made to take place by coupling with energy source: another reaction or energy source: another reaction or electric currentelectric current
ElectrolysisElectrolysis
2NaCl(l) => 2Na(s) + 2Cl2(g)